Re: exponential equation with constant



achava@xxxxxxxxxxx wrote:
> Let's see. Using your trick of letting y = exp(-ct), assuming c != 0,
> yields
>
> b*(y^(d/c)) + a*y - 1 = 0.
>
> The means of solution will now depend on the nature of d/c. If it
> happens to be 2, you have a quadratic equation and are good to go. If
> it is 3 or 4 there are exact solutions. Otherwise you need, not a
> computer simulation, but an iterative algorithm that will solve the
> equation. These should be pretty easy to come up with, or else you
> could just use the good old Newton-Raphson.

You could also use Newton-Raphson on the original:

f(t) = 1 - b*exp(-dt) - a*exp(-ct)
f'(t) = 1 + bd*exp(-dt) + ac*exp(-ct)

Iteration: t = t - f(t)/f'(t)

Here's a numerical experiment starting at t=1 using
that iteration rule:

t =
0
t =
3.999200159968006e-04
t =
6.304508579213751e-04
t =
6.899299967322885e-04
t =
6.931364420750211e-04
t =
6.931471733098213e-04
t =
6.931471805551151e-04

- Randy

.